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How do I generate a random MAC address from the Linux command line?

I search for a solution that only requires standard tools commonly found on the Linux command line.

The MAC address will be used for a guest KVM.

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5 Answers 5

up vote 24 down vote accepted

I use

macaddr=$(echo $FQDN|md5sum|sed 's/^\(..\)\(..\)\(..\)\(..\)\(..\).*$/02:\1:\2:\3:\4:\5/')

The benefit of this method, over a completely random number, is that it's possible to reliably reproduce the MAC address based on the FQDN of the machine, which I find useful sometimes. The 02 for the first octet just sets the "locally assigned" bit, which makes it obvious that it's not a vendor-provided MAC address, and guarantees that you won't collide with a real NIC's MAC address.

If you need to generate multiple MAC addresses per host, I used to concatenate the FQDN with the name of the bridge to connect the interface to; this did a good job of spreading things out for different NICs.

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+1 for the reproducibility; for certain applications, that makes it a much superior method to mine. –  MadHatter Aug 10 '11 at 8:18
    
Thanks a lot, I like the idea of having it reproducible. –  Erik Sjölund Aug 10 '11 at 11:58
    
Plus no mac address conflicts in the unlikely event you'd randomly generate the same mac twice –  Petter H Sep 25 at 8:06

I know this post is old, but for future visitors, if you want a cryptographically secure pseudorandom MAC address, without being limited to 0x02 as the OUI, here is a fast mostly platform agnostic generator:

$ printf '%02x' $((0x$(od /dev/urandom -N1 -t x1 -An | cut -c 2-) & 0xFE | 0x02)); od /dev/urandom -N5 -t x1 -An | sed 's/ /:/g'
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Here's another one, based on wombie's answer:

macaddr=$(dd if=/dev/urandom bs=1024 count=1 2>/dev/null|md5sum|sed 's/^\(..\)\(..\)\(..\)\(..\)\(..\)\(..\).*$/\1:\2:\3:\4:\5:\6/')
echo $macaddr
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There's no need to run urandom output through md5sum; you can just use od as per Aaron Toponce's answer. –  womble Sep 25 at 23:29

The posted scripts are good, but I want to add a warning: Mind the Birthday (paradoxon)!

It comes from the fact that even if you have just 23 people, the chance is already 50% that 2 of them have birthday on the same day.

It depends on your scenario how you use it, but if you generate the MACS randomly, at approx 1 million your chance for a mac number clash is 40% at 2 million it is already 87%!

If you need just a couple this is ok, but when you maintain a server farm with hundreds of servers, each of them hosting tens of virtual machines, or if you use the macs as index in some db for bookkeeping and you need uniques be careful!

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Thanks, for the warning about the Birthday paradox! In my case I will take the risk as I will generate around 20 MAC addresses. –  Erik Sjölund Aug 10 '11 at 12:22
1  
If you're running hundreds of servers each hosting tens of virtual machines all on the same broadcast domain, you've got bigger problems than MAC address collision risk. –  womble Sep 25 at 23:28
myserver% perl -e 'for ($i=0;$i<6;$i++){@m[$i]=int(rand(256));} printf "%X:%X:%X:%X:%X:%X\n",@m;'
55:C2:A5:FA:17:74

Ah, the ol' Swiss Army Chainsaw rides again. And by way of version 0.2, I'm unashamedly stealing womble's excellent point about the first octet being 02:

myserver% perl -e 'for ($i=0;$i<5;$i++){@m[$i]=int(rand(256));} printf "02:%X:%X:%X:%X:%X\n",@m;'
02:8E:94:A3:47:26
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Thanks MadHatter, I tried your second variant and it worked. Very nice! –  Erik Sjölund Aug 11 '11 at 12:47

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